08 - Method of Characteristics for First-Order Semi-Linear PDE

Method of Characteristics

We have First Order Semi-Linear PDE with boundary data

Suppose we are seeking solution $z$ along curve ${\gamma }_0=({\gamma }_1,{\gamma }_2)$ (data curve) in the $(x,y)$-plane

So we have $z(x,y)=g(x,y)$ for some function $g$ for all points on data curve ${\gamma }_0$
By setting ${\gamma }_3=g({\gamma }_1,{\gamma }_2)$ we have curve $\gamma =({\gamma }_1,{\gamma }_2,{\gamma }_3)$ which is called the initial curve
where the initial curve $\gamma$ must be in the solution surface $\Sigma$

So we have the following steps (method of characteristics)

1. Parametrise $\gamma$ over some interval $I$

$$\gamma (s)=({\gamma }_1(s),{\gamma }_2(s),{\gamma }_3(s))\text{ for }s\in I$$

Assume that each component of $\gamma$ is continuously differentiable

2. Determine the solutions $(x(t,s),y(t,s),z(t,s))$ of characteristic equations

$$\begin{array}{rl}\dot{x}& =P(x,y)\\ \dot{y}& =Q(x,y)\\ \dot{z}& =R(x,y,z)\end{array}$$

with initial data $x(0,s)={\gamma }_1(s)$, $y(0,s)={\gamma }_2(s)$, $z(0,s)={\gamma }_3(s)$ for $s\in I$

3. This gives us solution surface $\Sigma$ in parametric form

$$\Sigma =\{(x(t,s),y(t,s),z(t,s))\}$$

for all $s\in I$
For each $s$, consider the maximal set of $t$‘s which solves all three characteristic equations

4. Eliminate the parameters $s,t$ and write $\Sigma$ as a graph to read off the solution
   Note that there is a restriction on the data for method to work

Finding the Domain of Definition

If the initial curve is bounded then

The domain of definition is usually bounded by the projections of the characteristics through end points of the initial curve

See lecture notes page $55$ for an example

Change of Variables of the Surface $\text{S}$

Using the Jacobian

$$J(s,t)=\frac{\partial (x,y)}{\partial (t,s)}=\det \left(\begin{array}{cc}{x}_t& y_t\\ x_s& y_s\end{array}\right)$$

If

$$x=x(t,s)\text{ and }y=y(t,s)$$

are continuously differentiable functions of $t$ and $s$ in a neighbourhood of a point then

There exist

$$t=t(x,y)\text{ and }s=s(x,y)$$

which are unique and continuously differentiable functions
in some neighbourhood of the point with $J$ being non-zero at that point

By substituting it back to $z=z(t,s)$ we get

$$z=z(t(x,y),s(x,y))=f(x,y)$$

where $z$ is continuously differentiable function of $x$ and $y$

General Solution of of an PDE

Expect the most general solution of a $n$-order ODE is to have $n$-arbitrary constants
Hence we expect the most general solution of a PDE of order $n$ to have $n$ arbitrary functions